如何检索员工列表'；SQL中的超级用户

基本上，我有一个表格，列出了员工的编号和他们向谁汇报的员工编号

 EmpNum | Name         | ReportsTo
 ---------------------------------
 1234   | John Smith   | 4523
 3245   | Annie Apples | 1234
 1532   | Bob Rogers   | 3245
 6574   | Dong Wong    | 1532

等等等等。所以黄东的等级结构是：他向->1532汇报，哪个向->3245汇报，哪个向->1234汇报

（我是SQL新手，所以希望能有清晰易懂的解决方案）

通过加入

select e1.EmpNum, e1.name, e2.EmpNum as BossNum, e2.name as BossName from empTable e1 join empTable e2 on e1.ReportsTo=e2.EmpNum

---

连接可以是您想要的任意长度，e3、e4。。。但是在前端以编程方式进行可能更容易。

您没有指定DBMS，因此这是标准的ANSI SQL

with recursive report_tree as (
  select empnum, name, reportsto
  from employees
  where empnum = 6574 
  union all
  select c.empnum, c.name, c.reportsto
  from employees c 
    join report_tree p on p.reportsto = c.empnum
) 
select *
from report_tree;

如果您想要“图形化”显示，可以这样做（仍然是标准的ANSI SQL）：

---

试试这个，cte最适合递归。所以对于这个问题已经有了很多解决方案

create table #emp(
EmpNum int,
Name varchar(50),
ReportsTo int
);

insert into  #emp
values
(1234,'John',4523),
(3245,'Annie',1234),
(1532,'Bob',3245),
(6574,'Dong',1532)



with rec as (
    select #emp.ReportsTo, #emp.EmpNum, #emp.Name, 1 as level from #emp where Name = 'Bob'
    union all
    select  #emp.ReportsTo, #emp.EmpNum, #emp.Name, level + 1 as level from #emp    
    inner join rec   
    on #emp.EmpNum = rec.ReportsTo
)
select ReportsTo, EmpNum, Name, level from rec
where level = (select max(level) from rec)
OPTION (MAXRECURSION 0)

---

你用的是哪种数据库管理系统？你能证明你自己在努力解决这个问题吗？嗯，我意识到你需要一个完整的链。。。有多少层？或者这可能是无限的？

create table #emp(
EmpNum int,
Name varchar(50),
ReportsTo int
);

insert into  #emp
values
(1234,'John',4523),
(3245,'Annie',1234),
(1532,'Bob',3245),
(6574,'Dong',1532)



with rec as (
    select #emp.ReportsTo, #emp.EmpNum, #emp.Name, 1 as level from #emp where Name = 'Bob'
    union all
    select  #emp.ReportsTo, #emp.EmpNum, #emp.Name, level + 1 as level from #emp    
    inner join rec   
    on #emp.EmpNum = rec.ReportsTo
)
select ReportsTo, EmpNum, Name, level from rec
where level = (select max(level) from rec)
OPTION (MAXRECURSION 0)